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P->Q

  • 2 years ago

simplify into a single fraction and write your answer in terms of sin x and cos x: tan x + 1/tan x

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  1. ZeHanz
    • 2 years ago
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    Hint: tanx=sinx/cosx, so 1/tanx=cosx/sinx. You now only need to write the sum of these as one fraction!

  2. P->Q
    • 2 years ago
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    so sin^2x + Cos^2x/cosxsinx ?

  3. P->Q
    • 2 years ago
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    anyone?

  4. asnaseer
    • 2 years ago
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    yes - that is correct but you can simplify this further.\[\sin^2(x)+\cos^2(x)=?\]

  5. P->Q
    • 2 years ago
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    o um (sinx)(sinx) + (cosx)(cosx)/cosxsinx

  6. asnaseer
    • 2 years ago
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    no, the equation I wrote up there is a well know trig identity - look it up in your notes, you must have been taught this at some stage.

  7. P->Q
    • 2 years ago
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    1

  8. P->Q
    • 2 years ago
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    Pythagorean trig identity

  9. asnaseer
    • 2 years ago
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    correct

  10. asnaseer
    • 2 years ago
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    so now make use of that fact to simplify your expression

  11. P->Q
    • 2 years ago
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    hmmm

  12. P->Q
    • 2 years ago
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    ya idk, I just kinda put it out sin^2x + Cos^2x/cosxsinx = 1 im not seeing how this relates it just made it worse,.

  13. asnaseer
    • 2 years ago
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    the identity you need to use is:\[\sin^2(x)+\cos^2(x)=1\]

  14. P->Q
    • 2 years ago
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    what do u mean, this has nothing to do with this problem lol I mean all I know is that is the Pythagorean tri identity and that you can move cos or sin and lets say u move sin it goes to the right and u have cos^2(x) = plus or minus radical 1-sin^2(x)

  15. asnaseer
    • 2 years ago
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    Look at the steps you took in this problem:\[\begin{align} \tan(x)+\frac{1}{\tan(x)}&=\frac{\sin(x)}{\cos(x)}+\frac{\cos(x)}{\sin(x)}\\ &=\frac{\sin^2(x)+\cos^2(x)}{\cos(x)\sin(x)} \end{align}\]now notice that the numerator contains the terms: \(\sin^2(x)+\cos^2(x)\), and you know that this identity always equals 1. Therefore you can replace \(\sin^2(x)+\cos^2(x)\) by 1.

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